Chapter 4 Quantum entanglement · Quantum entanglement ... interact indirectly through their direct...

Chapter 4

Quantum entanglement

In this chapter we study the nature of a special type of correlation displayedby the entangled states. These correlations have no classical counterpart, inother words, they cannot be described by classical probability distributions.They are genuine quantum mechanical correlations built up in the states ofcomposite quantum systems.

We first take a close look at the so-called Bell states which violate thefamous Bell inequalities1. These states display the essence of entanglementand the CHSH inequality provides an experimentally testable signature ofit. We then describe three applications: a quantum key distribution protocol(Ekert 1991), quantum teleportation and dense coding. We stress here thatall three of them have been experimentally realized, and form importantprimitive protocols for quantum communication.

In quantum information processing one tries to use entanglement as aquantifiable resource, much like energy or information, and it would be veryconvenient to be able to measure the degree or quantity of entanglement.Finding such a measure is however non-trivial. We will come back to thispoint in later chapters.

4.1 Bell states

Production of Bell states. We have seen in chapter 2 that in order toproduce entangled states the Qbits must “interact”, at some point in time.The prototypical example of entangled states are the Bell states which form

1There is a class of such inequalities named after John Bell who derived the firstones. In this chapter we derive the more transparent Clauser-Horne-Shimony-Holt (CHSH)inequality.

1

2 CHAPTER 4. QUANTUM ENTANGLEMENT

H|x>

|y>

|Bxy>|

Figure 4.1: Quantum circuit producing Bell states.

a basis of C2 ⊗ C2. These can be produced from the unitary gate

U = (CNOT )(H ⊗ I) (4.1)

This is a 4 × 4 matrix equal to the usual matrix product of the two 4 × 4matrices CNOT and H ⊗ I. The Control Not gate provides the interactionbetween the two bits. It is defined as the NOT gate acting on the second bitprovided the first one2 equals 1

CNOT |x, y〉 = |x, y ⊕ x〉

The matrices H and I are the usual 2 × 2 Hadamard and identity matri-ces. The circuit representation of the unitary gate U = (CNOT )(H ⊗ I) isdepicted in figure 1.

Let us calculate the action of this circuit on a tensor product state |x〉 ⊗|y〉 = |x, y〉.

(CNOT )(H ⊗ I)|x, y〉 = (CNOT )1√2

(|0〉+ (−1)x|1〉

)⊗ |y〉

=1√2CNOT |0, y〉+

(−1)x

√2

CNOT |1, y〉

=1√2|0, y〉+

(−1)x

√2|1, y ⊕ 1〉)

= |Bxy〉

More explicitly we have

|B00〉 =1√2

(|00〉+ |11〉

)= U |00〉

|B01〉 =1√2

(|01〉+ |10〉

)= U |01〉

2called the control bit

4.1. BELL STATES 3

|B10〉 =1√2

(|00〉 − |11〉

)= U |10〉

|B11〉 =1√2

(|01〉 − |10〉

)= U |11〉

These four states are a unitary “rotation” of the four canonical basis statesof C2 ⊗ C2 and thus also form a basis, called the Bell basis.

Here the interaction is effected by the CNOT gate: building such a gatein a laboratory requires bringing two particles supporting the Qbits |x〉 and|y〉 close enough in space and time (interactions are local). Photons do notinteract directly with one another (Maxwell equations are linear) but they caninteract indirectly through their direct interaction with matter (one speaksof non-linear optics). Localized sources producing pairs of entangled photonsare excited atoms or nuclei, emitting photons when they fall in their groundstate. Electron spin can also be entangled because the combination of theCoulomb interaction with the Pauli principle can produce special magneticcorrelations. In fact this kind of entanglement is very common place: in ahydrogen molecule the spin part of the chemical valence bond*3 between twohydrogen atoms is the state

|B11〉 =1√2(| ↑↓〉 − | ↓↑〉)

The reader should check that

|B00〉 =1√2

(|00〉+ |11〉

)=

1√2

(|γγ〉+ |γ⊥γ⊥〉

)(4.2)

where |γ〉 is any state of C2. This has remarkable consequences as the follow-ing discussion will show. For the sake of the argument we suppose that Alicehas captured one photon in her lab and Bob has captured the other photonin his lab (figure 2). Irrespective how remote the two labs are, it is alwaystrue that the two photons have come from a common localized source. Nowwe look at the outcome of several simple measurements that Alice and Bobmight do each in their own lab. We are assuming that they cannot communi-cate the outcomes of these measurements. We will consider the three specificsituations where: Alice measures first/Bob measures after; Bob measuresfirst/Alice measures after; Alice and Bob measure simultaneously4.

3the anti-symmetry of the spin part allows the orbital part to be in the symmetricenergetically favorable state (Heitler-London theory)

4For definiteness we have in mind a Galilean picture of space-time. However the dis-cussion is essentially the same for a relativistic picture for space-time.


|B>

ALICE

BOB

Figure 4.2: Alice and Bob share an entangled pair.

• Alice measures first and Bob after. The “measurement apparatus” ofAlice is formed by the projectors {|α〉〈α| ⊗ I, |α⊥〉〈α⊥| ⊗ I} so that,according to the measurement postulate, the Bell state collapses on oneof the projections (remember we have to normalize after projecting)

|α〉〈α| ⊗ I|B00〉 =1√2|αα〉 → |α〉 ⊗ |α〉, with prob

1

2

|α⊥〉〈α⊥| ⊗ I|B00〉 =1√2|α⊥α⊥〉 → |α⊥〉 ⊗ α⊥〉, with prob

1

2

Therefore Alice observes her photon in the collapsed state |α〉 or |α⊥〉.Bob, on his side, does not know anything, and doesn’t even know thatAlice has performed measurements ! In order for him to learn some-thing he can try to perform a measurement on his photon. But he hasto choose a basis {|β〉, |β⊥〉}. Given that his photon is in the state |α〉,his photon collapses to |β〉 with prob cos2(α− β) or to |β⊥〉 with probsin2(α − β). Similarly, given that his photon is in the state α⊥ we getthe same result with cos2 and sin2 interchanged. The fact that Bobdoes not know the initial state of his photon or that he does not evenknown what Alice has done should not bother you: the point is that hedoes a specific experiment (measurement in the β, β⊥ basis) and findsa net outcome. The net outcome in Bob’s lab is that the photon is inthe state |β〉 with prob 1

2or |β⊥〉 with prob 1

2.

• Bob measures first and Alice after. The same discussion shows that, ifBob performs measurements first (in the β, β⊥ basis) while Alice sleepsand Alice measures after (in the (α, α⊥ basis) the net outcome of eachparty is the same.

• Bob and Alice measure simultaneously. You might think (?) that ifboth parties perform simultaneous local measurements the whole sce-

4.1. BELL STATES 5

nario is different. Let us try. Suppose Alice and Bob perform simulta-neous measurements in the basis

{|α, β〉, |α, β⊥〉, |α⊥, β〉, |α⊥, β⊥〉}

The Bell state

|B00〉 =1√2(|00〉+ |11〉) =

1√2(|γγ〉+ |γ⊥γ⊥〉)

will collapse to one of the four basis states. So Alice will be in possessionof a photon in state |α〉 or |α⊥〉 and Bob in possession of a photon inthe state |β〉 or |β⊥〉. The situation is exactly the same than in theprevious situations ! It is very instructive to compute the probabilitiesof the respective collapsed states (which are nothing else than the basisstates). One finds that these are5

1

2cos2(α− β),

1

2sin2(α− β),

1

2sin2(α− β),

1

2cos2(α− β)

Alice finds that the probability of her outcomes |α〉 (resp |α⊥)

1

2cos2(α− β)

1

2sin2(α− β) =

1

2

(for both cases) as in the previous scenarios; and the same holds truefor Bob. Therefore the conclusions that Alice and Bob infer fromtheir simultaneous local measurements are the same than in the non-simultaneous cases above.

To summarize the situation, we see that when Alice and/or Bob performsuccessive or simultaneous local measurements on their photons, whatever istheir choice of basis they find the photon in one of the two chosen basis stateswith probability 1

2. In other words the entropy of the probability distribution

of their local outcomes is maximal (it equals ln 2 bits) and they may infer thattheir photon is in a “maximally disordered state“. In fact if they don’t knowthat the source produced an entangled pair or if nobody tells them that thetwo photons are entangled they have no way of even noticing that the pairis entangled. It seems that we have no way of knowing if we are entangledto some distant Alien in the universe, just by performing local experimentsin our part of the universe (scary no ?!). We will see in the next section thatAlice and Bob can assert that their photons are entangled if they are allowedto communicate. Here by communicate we mean the perfect or approximatetransmission of a message.

5fortunately independent of γ


A B

space

time

space

time MINKOWSKIGALILEE

A

B

A

B

A B

Figure 4.3: Galilean space-time: B is in the future of A (left); B and A aresimultaneous (right). Minkowski space-time: the slopes of the light-cones equal1/c (in natural relativistic units the speed of light is c = 1 as on the picture). B isin the future light cone of A (left); B and A are space-like separated and cannotbe causally related. Note that for c → +∞ the slope of the light cone vanishesand one recovers the Galilean picture.

Let us also point out that here we have discussed the situation havingin mind a Galilean picture of space-time. In other words the meaning ofthe words ”before”, ”simultaneous” and ”after” is the ”usual” one. Howeverthis is only an approximation and one might question if a proper account ofMinkowskian space-time (see figure 4.3) would change our conclusions. Ac-cording to the theory of special relativity these words are relative to each ob-server’s frame of reference. What has an absolute meaning is the space-timeinterval which may be space-like, time-like (or zero). If the local measurementevents (events are points in space-time) of Alice and Bob are separated bya space-like vector there cannot possibly be a causal connection between theevents, and in particular it is guaranteed that Alice and Bob cannot establisha classical communication link during the experiment. On the other hand ifthe measurement events are separated by a time-like vector it is conceivablethat there is a causal connection between the events, however unless Aliceand Bob set up such a communication link, there is no reason to believe thatthere is a causal connection between the outcomes since they are exactly thesame as in the case of space-like separation.

4.2 Bell inequalities and Aspect experiment

We saw in the last section that if there is no communication between Aliceand Bob they can only infer that the photons are in a maximally disorderedstate. In this section we will see that by doing repeated measurements andby communicating the results afterwards, Alice and Bob can assert if the

4.2. BELL INEQUALITIES AND ASPECT EXPERIMENT 7

ALICE BOB

|B>

Figure 4.4: Experimental set up

state produced by the source is entangled or not.

The procedure that we are going to describe was initially invented byJohn Bell, and motivated by a famous paper of Einstein-Podolsky-Rosen.The later claimed that the entangled states do not provide a ”complete“description of the correlations present in the system, and where seeking a”classical“ theory of these correlations. Bell’s approach to the problem is totry to decide if the correlations in a real pair of entangled photons (producedby an excited atomic source say) can be described or cannot be described by aclassical theory. The general idea is that if a pair of photons is described by aclassical theory then appropriate correlation functions of the measurements ofAlice and Bob satisfy very special constraints. These constraints are violatedif the pair is described by a quantum mechanical Bell state. We will see thatBell’s approach is able to discriminate between a huge set of classical theoriesand QM. Famous experiments of Aspect-Grangier-Roger have shown thatstandard QM wins!

The experimental protocol. A source S produces, at each instant of timen, a pair of photons. We do not have any prejudice as to what is the stateor the description of the pair. One photon flies to Alice’s lab and the otherflies to Bob’s lab. In each lab our two protagonists operate independently:they do not communicate and do not care what the other one does.

• At each time instant n, Alice randomly uses analyzers

{|α〉, |α⊥〉} or {|α′〉, |α′⊥〉}

to measure the polarization of her photon. When she records a click inthe detector she sets an = +1 or a′n = +1 and when the detector doesnot click she sets an = −1 or a′n = −1. She keeps track of her choicesfor the analyzer at each n.


• At each time instant n, Bob randomly uses analyzers

{|β〉, |β⊥〉} or {|β′〉, |β′⊥〉}

to measure the polarization of his photon. When he records a click inthe detector he sets bn = +1 or b′n = 1 and when the detector doesnot click he sets bn = −1 or b′n = −1. He keeps track of his choices ofanalyzers for each n.

• Now there is a classical communication phase. Alice and Bob meetand discuss all their measurements. They classify them according tothe four experimental setups. At each time instant n the possible ar-rangements of analyzers were

1 = (α, β), 2 = (α, β′), 3 = (α′, β), 4 = (α′, β′)

For each arrangement they compute the following empirical averages

1

N1

∑n1

an1bn1 ,1

N2

∑n2

an2b′n2

,1

N3

∑n1

a′n3bn3 ,

1

N4

∑n4

a′n4b′n4

Then they compute the following correlation function

Xexp =1

N1

∑n1

an1bn1 +1

N2

∑n2

an2b′n2− 1

N3

∑n1

a′n3bn3 +

1

N4

∑n4

a′n4b′n4

Prediction of classical theories. We assume that the quantities that Aliceand Bob measure correspond to well defined observables A, A′, B, B′ thathave simultaneous definite values a, a′, b, b′ even when there is no observer(i.e. even before the measurement). This is the hypothesis of ”realism“and is discussed in more detail later. Basically, this is like saying that aparticle has a definite position and velocity (here analogous to a and a′) evenwhen these quantities are not observed or measured. Furthermore we assumethat the outcomes of Alice and Bob can be modeled by a joint probabilitydistribution6

Pclass(a, a′, b, b′)

Here by a, b, a′ and b′ we mean the random variables modeling the measure-ment outcomes. The expectation with respect to Pclass is denoted by Eclass.The corresponding theoretical prediction for each empirical average above is

Eclass[ab], Eclass[ab′], Eclass[a′b], Eclass(a

′b′)

6this second assumption follows from the assumption of ”realism” combined with ”lo-cality“. This is explained in the next paragraph


and using only the linearity of expectation

Xclass = Eclass[ab + ab′ − a′b + a′b′]

Notice that

ab + ab′ − a′b + a′b′ = a(b + b′) + a′(b′ − b)

and that

−2 ≤ a(b + b′) + a′(b′ − b) ≤ 2

Indeed if b = b′ then only the first term survives which leads to the inequal-ity; while if b 6= b′ only the second term survives which again leads to theinequality. Thus we have for the expectation,

−2 ≤ Xclass ≤ 2

This is one of the simplest Bell type inequalities which was derived byClauser-Horne-Shimony-Holt and is called the CHSH inequality.

In order to derive this result we haven’t assumed anything about thestate of preparation of the source. We have only assumed that the experi-mental results can be cast into a joint probability distribution. In fact this isnot a priory so obvious. There are four experimental arrangements so thatwhen Alice and Bob meet they have four histograms that can be fitted to 4probability distributions:

P1(a, b), P2(a′, b), P3(a, b′), P4(a

′, b′)

Are these the marginals of a common Pclass(a, a′, b, b′) ? It is not a prioryclear that, in this experiment, nature gives us histograms that are marginalsof a common joint distribution. In fact this is not always the case. Indeedany of us can construct four probability distributions that are not marginalsof a common one, and this is an outcome of our brains (viewed as a physicalsystems). So why is the assumption leading to the CHSH inequality veryreasonable ? We answer this question below, but do not attempt to providethe most general argument.

Let us admit that the laws of physics are ”local”. By this we mean thatwhen Alice (resp. Bob) perform measurements that are space-like separatedAlice’s experimental outcomes (resp. Bob’s) depend only on her own localchoice of analyzers. As far as we know, this is an assumption that underliesall the known (i.e. experimentally verified) fundamental laws of physics.

Furthermore let us suppose, following our classical intuition, or followingEinstein, that the outcomes of experiments should be well defined preexisting


functions7 of the system state and the experimental set-up. This is sometimescalled ”realism“.

In mathematical terms ”local realism“ means that there should be a func-tion, such that

a = fA(α; λ), a′ = fA(α′; λ), b = fB(β; λ), b′ = fB(β′; λ)

Here λ is a set of variables accounting for the state of the system and whateveris needed to compute the experimental outcome. It has become customaryto call them ”hidden variables”.

The hidden variables may be random or deterministic8 and their set ofvalues is described by a probability distribution h(λ). According to ”localrealism“ the histograms of Alice and Bob are modeled by

P1(a, b) =

∫dλh(λ)δ(a− fA(α, λ))δ(b− fB(β, λ))

P2(a, b′) =

∫dλh(λ)δ(a− fA(α, λ))δ(b′ − fB(β

′, λ))

P3(a′, b) =

∫dλh(λ)δ(a′ − fA(α′, λ)δ(b− fB(β, λ))

P4(a′, b′) =

∫dλh(λ)δ(a′ − fA(α′, λ)δ(b′ − fB(β

′, λ))

Evidently these are the marginals of a joint probability distribution

Pclass(a, a′, b, b′) =

∫dλh(λ)δ(a−f(α, λ))δ(a′−f(α′, λ))δ(b−f(β, λ))δ(b′−f(β, λ))

Prediction of QM for a Bell state. First of all we notice that accordingto the quantum formalism the measurements of Alice and Bob are measure-ments of the 4 observables (hermitian matrices)

A = (+1)|α〉〈α|+ (−1)|α⊥〉〈α⊥|, A′ = (+1)|α′〉〈α′|+ (−1)|α′⊥〉〈α′⊥|

and

B = (+1)|β〉〈β|+ (−1)|β⊥〉〈β⊥|, B′ = (+1)|β′〉〈β′|+ (−1)|β′⊥〉〈β′⊥|7we could also frame the discussion in a slightly more general context where the outcome

is described by a probability distribution pA(a|α, λ). Here we have pA(a|α, λ) = δ(a −fA(α, λ). The conclusions are however the same, but this remark is interesting because itshows that it is not determinism that is at stake here.

8in this case the distribution is simply a Dirac h(λ) = δ(λ− λ0)


At each time instant n the state of the photon pair is described by some ket|Ψ〉 ∈ C2 ⊗ C2. The quantum mechanical prediction for the four empiricalaverages of Alice and Bob is

〈Ψ|A⊗B|Ψ〉, 〈Ψ|A⊗B′|Ψ〉, 〈Ψ|A′ ⊗B|Ψ〉, 〈Ψ|A′ ⊗B′|Ψ〉

and for the correlation function

XQM = 〈Ψ|A⊗B|Ψ〉+ 〈Ψ|A⊗B′|Ψ〉 − 〈Ψ|A′ ⊗B|Ψ〉+ 〈Ψ|A′ ⊗B′|Ψ〉

Now let us compute this quantity for the Bell state

|Ψ〉 = |B00〉

The first average is best computed by expressing the Bell state as 1√2(|αα〉+

|α⊥α⊥〉).

〈B00|A⊗B|B00〉 =1

2〈αα|A⊗B|αα〉+

1

2〈α⊥α⊥|A⊗B|α⊥α⊥〉

+1

2〈αα|A⊗B|α⊥α⊥〉+

1

2〈α⊥α⊥|A⊗B|αα〉

=1

2〈α|A|α〉〈α|B|α〉+

1

2〈α⊥|A|α⊥〉〈α⊥|B|α⊥〉

=1

2· 1 ·

(|〈α|β〉|2 − |〈α|β⊥〉|2

)+

1

2· (−1) ·

(|〈α⊥|β〉|2 − |〈α⊥|β⊥〉|2

)=

1

2

(cos2(α− β)− sin2(α− β)

)− 1

2

(sin2(α− β)− cos2(α− β)

)= cos2(α− β)− sin2(α− β) = cos 2(α− β)

Performing similar calculations for the other averages we find

XQM = cos 2(α− β) + cos 2(α− β′)− cos 2(α′ − β) + cos 2(α′ − β′)

This quantity is maximized for the following choice of angles (and all globalrotations of this choice of course, figure 4),

α = 0, α′ = −π

4, β =

π

8, β′ = −π

8

and equals

XQM = cosπ

4+ cos

π

4− cos

3π

4+ cos

π

4= 2

√2

We see that the CHSH inequality is violated ! For the three other Bell stateson finds the same result. In the exercises you will show that this is the


a’

b’

a

b = pi/8

= 0

= −pi/8

= − pi/4

Figure 4.5: Optimal choice of analyzer orientation

maximum possible violation over all quantum states of C2 ⊗ C2. In thissense the Bell states are maximally entangled.

QM predicts that the four histograms of Bob and Alice are

P1(a, b) =1

4(1 + ab cos 2(α− β))

P2(a, b′) =1

4(1 + ab′ cos 2(α− β′))

P3(a′, b) =

1

4(1 + a′b cos 2(α′ − β))

P4(a′, b′) =

1

4(1 + a′b′ cos 2(α′ − β′))

For example: P2(+1,−1) = |〈α, β′⊥|B00〉|2 = 14(1 − cos 2(α − β)). There are

special choices of the angles α, β, α′, β′ for which these are not the marginalsof a common distribution Pclass(a, b, a′, b′) otherwise we would have |X| ≤ 2:this is just a mathematical fact9. Now, nature produces these four histogramsin an experiment satisfying locality in the sense that all analyzer choices ofAlice and Bob are independent. But she plays a very subtle magic trick withus: the correlations that are built up in Bell’s states are non-local in the sensethat correlations are present in the measurement outcomes even though themeasurements on the photons are purely local. Alice and Bob cannot noticethese non local correlations by purely local means in their own lab. Theyhave to meet or to communicate by exchanging matter.

Experiments. In a famous set of experiments performed in the 80’s Aspect-Grangier-Roger showed that experiment agrees with QM and not with clas-sical theories. The difficulty of these experiments is that, one wants to rotatethe analyzers of Alice and Bob fast enough so that the measurement events

9In some sense they are the marginals of a quantum state

4.3. EKERT PROTOCOL FOR QKD 13

are separated by a space-like interval. Otherwise, one may always argue thatsome form of classical communication or interaction conspires to make upthe results (on speaks of locality loophole). This is the challenge that theAspect experiments were the first to address, as compared with other slightlyearlier experiments. This locality loophole has been since then conclusivelysettled by more recent experiments10. There are other issues, that one hasto address in principle, such as the efficiency of coincident detections (calleddetection loophole). So far there are no experiments that completly addressall loopholes at the same time.

The Aspect experiments tell us that we have to abandon the ”local re-alism“. QM does not give up locality in the sense that fundamental modelsof interactions are local. Rather it gives up realism in the sense that e.g onecannot assert position and velocity (or a and a′) are predetermined beforea measurement. Indeed, when Alice performs a measurement her outcomedoes not depend on what Bob does (this is locality) but at the same time itis not a well defined function f(·, λ) independent of her choice of analyzers(in this sense realism doesn’t hold). There cannot exist such a function de-pending on hidden variables λ with distribution h(λ), which accounts for theexperimental results.

It is sometimes said that QM is non-local: this has to be understood in thesense that quantum mechanical states of the Hilbert space can be non-local, inother words arbitrarily extended; however the physical laws of interactionsare, as far as we know, local.

As you can begin to suspect it does not make much sense to stick to”classical intuitions“, and words like ”local realism“ or anything of this sort.We have to face new concepts and develop new intuitions.

4.3 Ekert protocol for QKD

A nice application of the CHSH inequality is a protocol for the generationof a secret key by two parties. We assume that a localized source of EPRparticles delivers entangled Qbits to Alice and Bob at each time instant n inthe state

|B00〉 =1

2(|00〉+ |11〉) =

1

2(|θθ〉+ |θ⊥θ⊥〉)

Moreover they have also established a noiseless communication channel.

The protocol:

10see the review by Anton Zeilinger ”Experiment and the foundations of quantumphysics“, in Reviews of Modern Physics 71, S288-S297 (1999)


a_3

a_2

a_1

=0

= −pi/8

= −pi/4

b_1=−pi/8

b_2=0

b_3=pi/8

Figure 4.6: Alice and Bob’s random choices of analyzers

• Alice has analyzers oriented in directions a1, a2, a3 and records theresults of measurements, at each time instant, for the observables

A(a) = (+1)|a〉〈a|+ (−1)|a⊥〉〈a⊥|

where she chooses a randomly among a1, a2, a3 (figure 5).

• Bob has three analyzers oriented along b1, b2, b3 and records theresults of measurements, at each time instant, for the observables

B(b) = (+1)|b〉〈b|+ (−1)|b⊥〉〈b⊥|

where he chooses b randomly among b1, b2, b3 (figure 5).

• Alice and Bob start a public discussion over the communication chan-nel: they inform each other on what vectors they used at each timeinstant.

• They do a security check to ensure that no eavesdropper is present.Alice and Bob select the time instants when the basis choices were

(a3,b3), (a3,b1), (a1,b1), (a1,b3)

Note that these are the same four analyzer arrangements used for theBell inequalities (figure 6). For such configurations and only for suchones they exchange their measurement results. Each party computesan empirical correlation coefficient

Xexp = Av[an(a3)bn(b3)] + Av[an(a3)bn(b1)]

− Av[an(a1)bn(b3)] + Av[an(a1)bn(b1)]

where Av is the empirical average. In a perfect world they should findXexp = 2

√2. We will see later that when an eavesdropper is present

4.3. EKERT PROTOCOL FOR QKD 15

b_3=pi/8

a_3=0

b_1=−pi/8

a_1=−pi/4

Figure 4.7: CHSH configuration

they will certainly find Xexp ≤ 2 because the effect of the eavesdropperis to destroy the entanglement of the EPR pair and the system thenbehaves ”classically“. The security check thus consists in checking that

Xexp > 2

If the test passes they conclude there is no eavesdropper and generatethe key, if not they stop communication.

• The key generation process is as follows. For every time n such thatthey used the same basis - that is (a3,b2) or (a2,b1) - they know forsure that

an = bn = 1, or an = bn = −1

(one can also check that in this case 〈B00|A⊗B|B00〉 = cos 2 ˆ(a,b) = 1).Thus they have a common subsequence of ±1’s that they keep secretand forms their shared secret key.

Attacks from Eve. Let us consider the simplest measurement attack inwhich Eve captures each photon of the EPR pair and makes a measurement(figure 7). Then she sends each photon (in the resulting state) to Alice andBob. She measures Alice’s photon in the basis {ea, e

⊥a } and Bob’s photon

in the basis {eb, e⊥b }. Her strategy for the successive choices of basis at each

time instant is described by a probability distribution

ρ(ea, eb) ≥ 0,

∫ ∫d2ead

2eb ρ(ea, eb) = 1

After Eve’s measurement the pair of photons is left in one of the fourtensor product states

|ea, eb〉, |ea, e⊥b 〉, |e⊥a , eb〉, |e⊥a , e⊥b 〉


ALICEBOB

EVE

|B>

e_a e_b

a_1, a_2, a_3 b_1,b_2,b_3

Figure 4.8: Eve collapses the pair in a tensor product state

with corresponding probabilities

|〈ea, eb|B00〉|2 =1

2cos2 (ea, eb), |〈ea, e

⊥b |B00〉|2 =

1

2sin2 (ea, eb)

|〈e⊥a , eb|B00〉|2 =1

2sin2 (ea, eb), |〈e⊥a , e⊥b |B00〉|2 =

1

2cos2 (ea, eb)

Let us compute the correlation coefficient that Alice and Bob would findduring the security test. Given Eve’s choice (ea, eb) we have

X(ea, eb) =1

2cos2 (ea, eb)S(ea, eb) +

1

2sin2 (ea, e⊥b )S(ea, e

⊥b )

+1

2sin2 (e⊥a , eb)S(e⊥a , eb) +

1

2cos2 (e⊥a , e⊥b )S(e⊥a , e⊥b )

where S(v,w) is the correlation coefficient for a pair of photons in the state|v,w〉 resulting from Eve’s measurement,

S(v,w) = 〈v,w|A(a3)⊗B(b3)+A(a3)⊗B(b1)−A(a1)⊗B(b3)+A(a1)⊗B(b1)|v,w〉

The average correlation coefficient found by Alice and Bob when Eve operatesis

X =

∫ ∫d2ead

2eb ρ(ea, eb)X(ea, eb)

We leave it as an exercise to check that |S(v,w)| ≤ 2. This is not be toosurprising since |v,w〉 is a tensor product state. This immediately leads to,

|X| ≤ 2.

Thus Alice and Bob notice the presence of Eve. Note that Eve could ma-nipulate (unitarily) the pair after her measurements in order to send otherphoton states to Alice and Bob. However if she re-entangles the photons she

4.4. QUANTUM TELEPORTATION 17

behaves as a new source for Alice and Bob, and she gets no information fromtheir measurements !

Finally let us note that if Eve copies the EPR pair (this can be done witha machine that copies the four orthogonal Bell states) and waits for the publicdiscussion before doing the measurements, she gets no information about thesecret key. Indeed her measurements operate on a different pair and thusshe get the same result than Alice and Bob only half of the time. This isequivalent to flip a coin at each time instant and cannot yield information.

Experiments. see in Review of Modern Physics 74 p 145-190 (2002) the ex-tensive article ”Quantum cryptography“ by N. Gisin, G. Ribordy, W. Tittel,H. Zbinden.

4.4 Quantum teleportation

Suppose that Alice and Bob are spatially separated and that Alice possessesa Qbit state,

|Φ〉 = α|0〉+ β|1〉, |α|2 + |β|2 = 1

The state (i.e α and β) is not necessarily known to Alice and is not knownto Bob. They also share an EPR pair

|B00〉 =1√2(|00〉+ |11〉)

and have at their disposal a classical communication channel.We are going to explain that by sending only two classical bits of infor-

mation over the classical channel, Alice can teleport the state to Bob. Hereteleportation means that |Φ〉 is destroyed in Alice’s lab and is reconstructedin Bob’s lab. Note that destruction of |Φ〉 in Alice’s lab is to be expected be-cause of the no-cloning theorem. After the teleportation process, Bob knowsthat he possesses the state |Φ〉 but still does not know the state itself (i.e hedoes not know α and β). We stress that the teleportation process involvesphysical transport of matter in the classical communication phase betweenAlice and Bob. Of course this classical communication phase cannot happenat speeds greater than that of light, so that the whole teleportation processdoes not violate the principles of relativity. We also note that the materialsupport of the state (e.g. photon polarization, electron spin) |Φ〉 = α|0〉+β|1〉is not necessarily the same in Alice’s and Bob’s lab.

Teleportation can be summarized by the following ”law“

teleporting 1 Qbit = sending 2 Cbits + sharing 1 EPR pair


and can be thought of, as some form of communication between Alice and Bobwhich share a classical channel and an ”EPR like channel“. The quantumstate |Φ〉 in Alice’s lab is erased on her side and reproduced in Bob’s lab -the information contained in α and β has not been communicated.

The protocol.

• A source produces an EPR pair of particles in the Bell state |B00〉23.One particle, called particle 2 is sent to Alice and one particle , calledparticle 3 is sent to Bob. The Hilbert space of the entangled system 23is H2 ⊗H3 = C2 ⊗ C2.

• Alice prepare a particle, called 1, in the state |Φ〉1 = α|0〉 + |β〉. TheHilbert space of particle 1 is H1 = C2.

• The total Hilbert space of the composite system 123 is H1⊗H2⊗H3 =C2 ⊗ C2 ⊗ C2 and the total state is

|Ψ〉 = |Φ〉1 ⊗ |B00〉23At this point a short calculation will facilitate the subsequent discussion

|Ψ〉 =α√2|000〉+

β√2|100〉+

α√2|011〉+

β√2|111〉

• Alice makes a local measurement in her lab, i.e on partiles 12. She usesan apparatus that has measurement basis of H1 ⊗H2

{|B00〉12, |B01〉12, |B10〉12, |B11〉12}

The associated projectors for the total system are

P00 = |B00〉〈B00|⊗I3, P01 = |B01〉〈B01|⊗I3, P10 = |B10〉〈B10|⊗I3, P11 = |B11〉〈B11|⊗I3

As usual the outcome of the measurement is one of the four possiblecollapsed states11 (check this calculation and also that the probabilityof each outcome is 1

4)

P00|Ψ〉 =1

2|B00〉12 ⊗ (α|0〉3 + β|1〉3)

P01|Ψ〉 =1

2|B01〉12 ⊗ (β|0〉3 + α|1〉3)

P10|Ψ〉 =1

2|B10〉12 ⊗ (α|0〉3 − β|1〉3)

P11|Ψ〉 =1

2|B11〉12 ⊗ (−β|0〉3 − α|1〉3)

11up to normalization

4.5. DENSE CODING 19

• Depending on the random outcome Bob has one of the four states

α|0〉3 + β|1〉3 = |Φ〉β|0〉3 + α|1〉3 = X|Φ〉α|0〉3 − β|1〉3 = Z|Φ〉β|0〉3 − α|1〉3 = iY |Φ〉

but he does not know the state he has.

• Alice knows that the outcome of the measurement (in her lab) is one ofthe four Bell states. She can thus use the Bell basis to re-measure (thiswill not perturb Bob’s particle this time) and determine her outcome.This outcome can be encoded by two classical bits

00, 01, 10, 11

that she sends to Bob over the classical communication channel. Assoon as Bob receives Alice’s message he knows that she has finishedher operations and he has the two bits of information needed to decidewhich unitary operation he has to perform on his state in order torecover |Φ〉,

I(α|0〉3 + β|1〉3) = |Φ〉X(β|0〉3 + α|1〉3) = |Φ〉Z(α|0〉3 − β|1〉3) = |Φ〉

−iY (β|0〉3 − α|1〉3) = |Φ〉

4.5 Dense coding

Suppose Alice and Bob have established a quantum channel over which theycan send Qbits (for example a optic fiber over which photons travel). Wewill study the capacity of such a noisy channel later in the course but forthe moment let us address a simpler question. Assume that Alice and Bobshare an EPR pair. How much information does one Qbit convey over thequantum channel ?

The answer is that 2 classical bits of information can be transmitted byAlice to Bob, by sending only 1 Qbit as long as they share an EPR pair. Theprotocol that achieves this is called dense coding.

We will come back to the problem of communicating classical/quantummessages over noisy quantum channels assisted/or not by entanglement in


later chapters. As we will see even for simple analogs of Shannon’s channelcoding theorem there are various open questions.

Dense coding can be summarized as follows:

communicating 2 Cbits = sending 1 Qbit + sharing 1 EPR pair

This ”law“ may seem complementary to the one of teleportation. Note how-ever that here only two particles are involved and it is the Qbit that isphysicaly transported form Alice to Bob.

Protocol.

• An EPR pair in the state |B00〉 is prepared by a source and each particlesent to Alice and Bob.

• Alice wants to communicate two bits of information to Bob:

– To send 00 she leaves her particle intact (or applies the unitarygate I) and physically sends her particle to Bob. Bob receives theparticle and is now in possession of the whole state

|B00〉

– To send 01 she applies the unitary gate X to her particle and thenphysically sends her particle to Bob. Bob is now in possession ofthe pair in the state

X1 ⊗ I2|B00〉 = |B01〉

– To send 10 she applies the unitary gate Z to her particle and thenphysically sends her particle. Bob is now in possession of the pairin the state

Z1 ⊗ I2|B00〉 = |B10〉

– To send 11 she applies the unitary gate iY to her particle andthen physically sends her particle. Bob is now in possession of thepair in the state

(iY )1 ⊗ I2|B00〉 = |B11〉

• Bob now has the EPR pair 12 in some state |Bxy〉. In order to determinethe two Cbits that Alice sent he must decide which Bell state he has.Since he knows that he has one of the four Bell states in his lab, he cando a local measurement in the Bell basis, and access the informationxy.

4.5. DENSE CODING 21

H H

Figure 4.9: Device for Bell basis measurements

Measurement in the Bell basis. One might think that measuring inthe Bell basis is a theoretician’s wishful thinking. In fact this has beenrealized experimentally, and although explaining how is beyond the scopeof this course, we give here an argument that shows that, in principle, itsuffices to have H and CNOT gates (the simplest unitary gates) togetherwith polarization analyzers (the simplest measurement apparatus).

We have seen at the beginning of this chapter that Bell states can begenerated as |Bxy〉 = (CNOT )(H ⊗ I)|xy〉. The projectors on the Bell basisstates are therefore related to the ones over the Z basis,

|Bxy〉〈Bxy| = (CNOT )(H ⊗ I)|xy〉〈xy|(H ⊗ I)(CNOT )

(here we have used that the Hadamard and control not matrices are her-mitian). The projectors |xy〉〈xy| correspond to the analyzer-photo-detectorapparatus for photons or to spin analyzers (Stern-Gerlach analyzer) for spins(Z basis). The circuit representation of a measurement device in the Bellbasis in given on figure 8. The input is any state |Ψ〉, and the output is oneof the four states

|Bxy〉〈Bxy|Ψ〉|〈Bxy|Ψ〉|

Experiments. Quantum teleportation and dense coding have been realizedexperimentally. A summary of the subject can be found in ”Les dossiers de larecherche“ no 18, fevrier 2005, ”L’etrange pouvoir de l’intrication quantique“,by N. Gisin.

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